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Question Number 143320 by mathdanisur last updated on 12/Jun/21

$$provethat:{tan}^{2}36°+{tan}^{2}72°=5$$

Answered by Ar Brandon last updated on 12/Jun/21

$${\tan }^{2}36°+{\tan }^{2}72°={\tan }^{2}36°+{\cot }^{2}18°$$$$={\left(\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}\right)}^2+{\left(\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}\right)}^2$$$$=\frac{10-2\sqrt{5}}{6+2\sqrt{5}}+\frac{10+2\sqrt{5}}{6-2\sqrt{5}}$$$$=\frac{(80-32\sqrt{5})+(80+32\sqrt{5})}{16}$$$$=\frac{160}{16}=10$$

Commented by mathdanisur last updated on 13/Jun/21

$$thanksSir..$$

Answered by MJS_new last updated on 13/Jun/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}$$

Answered by MJS_new last updated on 13/Jun/21

$${\tan }^{2}x+{\tan }^{2}2x=10$$$$\tan x=t$$$$\frac{t^2(t^4-2t^2+5)}{{(t^2+1)}^2}=10$$$$t^6-12t^4+25t^2-10=0$$$$(t^2-2)(t^4-10t^2+5)=0$$$$t=\pm \sqrt{2}\vee t=\pm \sqrt{5-2\sqrt{5}}\vee t=\pm \sqrt{5+2\sqrt{5}}$$$$t=\tan x\iff x=n\pi +\arctan t$$$$\mathrm{for}0⩽x<90°\mathrm{we}\mathrm{get}$$$$x=36°\vee x\approx 54.74°\vee x=72°$$

Commented by mathdanisur last updated on 13/Jun/21

$$Sirthatisitcannotbe5.?$$

Commented by MJS_new last updated on 13/Jun/21

$$\mathrm{no}.\mathrm{just}\mathrm{type}\mathrm{it}\mathrm{in}\mathrm{any}\mathrm{calculator}.$$

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